JAMB UTME - Physics - 2025

Bayanin Amsa

At constant pressure a fixed mass of gas obeys Charles' law: the volume is directly proportional to the absolute temperature, so

Converting the initial temperature to kelvin is the essential first step, because a ratio of Celsius temperatures is meaningless:

The gas shrinks to one-fifth of its original volume, so \(V_2 = \tfrac{1}{5}\times 2.00 = 0.40\,\mathrm{cm^{3}}\). Rearranging Charles' law:

Converting back to the Celsius scale asked for in the options:

Notice how the wording controls the arithmetic. Read as "the volume becomes one-fifth of the original", the volume ratio is \(1/5\) and the temperature falls by the same factor, giving \(-207\,^{\circ}\mathrm{C}\). Read instead as "the volume falls by one-fifth", the ratio would be \(4/5\) and the answer would be \(330\times0.8 = 264\,\mathrm{K} = -9\,^{\circ}\mathrm{C}\), which is not offered, so the first reading is the intended one.

The commonest error in this topic is to work in degrees Celsius, which here would give \(57/5 \approx 11\,^{\circ}\mathrm{C}\) and is completely wrong because the gas laws are proportionalities measured from absolute zero, not from the ice point. Always convert to kelvin before forming any ratio, and convert back only at the last line.

Bayanin Amsa

Thermal conductivity measures how readily a material passes heat on by conduction, that is by the transfer of energy from particle to particle without bulk movement of the material. Conduction depends on how closely and how strongly the particles are coupled, so it is best in solids (and outstanding in metals, where free electrons also carry energy), poorer in liquids, and worst in gases, whose molecules are far apart and rarely interact.

Air has by far the smallest value, so air is the poorest conductor of the four. This is exactly why insulating materials are designed to trap air rather than to be dense: cotton wool, fur, feathers, cavity walls and vacuum-flask jackets all work by holding air still. Ash insulates well for the same reason, but its own solid particles still conduct, so it cannot be a better insulator than the air within it.

A caution worth remembering: still air is a superb insulator, yet moving air carries heat away rapidly by convection. Conduction and convection are separate mechanisms, and a question about conductivity is asking only about the first. When the choices span different states of matter, rank them gas, liquid, non-metallic solid, metal in increasing order of conductivity and the answer usually follows at once.

Bayanin Amsa

For equilibrium, clockwise moment = anticlockwise moment about the pivot.

clockwise distance from pivot: 50cm - 10cm = 40cm

anticlockwise distance from pivot: 85cm - 10cm = 75cm

Applying the principle of moments

W x distance(w) = T x distance(T)

100 x 40 = T x 75

T = \(\frac{ 4000}{75}\) ? 53.33N

Bayanin Amsa

Net force in the horizontal (x) direction:
\(F_x = 8 \, \text{N} - 4 \, \text{N} = 4 \, \text{N} \quad \text{(to the right)}\)

Net force in the vertical (y) direction:
\(F_y = 15 \, \text{N} - 12 \, \text{N} = 3 \, \text{N} \quad \text{(3 N upward)}\)

Magnitude of the resultant force: \(R = \sqrt{F_x^2 + F_y^2} = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{N}\)

Bayanin Amsa

Every simple cell has three parts to identify separately: two electrodes, the electrolyte that conducts by ion movement between them, and often a depolariser that removes hydrogen gas from the positive electrode. The question asks only for the electrolyte, so the answer must be a substance that ionises in solution and carries charge inside the cell.

In the wet Leclanche cell the positive electrode is a carbon rod, the negative electrode is a zinc rod, and the electrolyte is a strong solution of ammonium chloride, \(\mathrm{NH_4Cl}\). It dissociates to give \(\mathrm{NH_4^+}\) and \(\mathrm{Cl^-}\) ions, which carry the current through the liquid while zinc dissolves at the negative electrode and hydrogen is released at the carbon rod. Manganese(IV) oxide, \(\mathrm{MnO_2}\), is packed round the carbon rod as the depolariser, oxidising the hydrogen to water and slowing down polarisation. The cell gives an e.m.f. of about 1.5 V but has a large internal resistance, so it suits work needing brief currents such as ringing a bell.

The other substances belong to different cells or to different parts of a cell. Carbon is the positive electrode of this same cell, which is why it is a tempting choice: an electrode is a conductor, not the ion-carrying solution. Lead(IV) oxide is the positive plate of the lead-acid accumulator, whose electrolyte is dilute sulphuric acid, and nickel hydroxide belongs to the alkaline nickel-cadmium or nickel-iron cell, whose electrolyte is potassium hydroxide. When revising cells, learn each one as a set of four labels: negative electrode, positive electrode, electrolyte, depolariser.

Bayanin Amsa

The key words here are very small. Measuring an ordinary resistance is one problem; measuring a resistance of a fraction of an ohm is a harder one, because the resistance of the connecting leads and of the sliding or soldered contacts is itself of that same order. Any method in which those stray resistances are counted along with the unknown will give a badly wrong result. The instrument that avoids this is the potentiometer.

In the potentiometer method the unknown low resistance \(R\) is joined in series with a known low standard resistance \(S\), so that exactly the same current \(I\) flows through both. The potential difference across each is then tapped off and balanced against a length of the potentiometer wire, giving balancing lengths \(l_1\) and \(l_2\). Since \(V = IR\) and the potentiometer reading is proportional to the potential difference,

Two features make this accurate for tiny resistances. At balance the galvanometer carries no current, so the potentiometer draws nothing from the circuit and does not disturb it, and the tappings are made directly across the resistance itself, so the lead and contact resistances lie outside the measured section and cancel out of the ratio.

A Wheatstone bridge, in the metre-bridge form, is the standard circuit for a moderate resistance of a few ohms upwards, and that familiarity is what makes it tempting here. It becomes unreliable at the extremes, however: for a very small unknown, the end corrections and the resistance of the jockey contact and connecting wires are comparable with the quantity being measured, so the balance point loses its meaning. A voltmeter is unsuitable because a real voltmeter draws some current from the circuit and the potential difference across a very small resistance is minute, so the reading would be dominated by instrument error. A rheostat is not a measuring instrument at all; it is a variable resistor used to control the current in a circuit.

The examination point to retain is that the range of the resistance decides the method: a bridge for middling values, and a potentiometer, whose null reading excludes lead and contact resistance, for very small ones.

Bayanin Amsa

Electromagnetic induction is described by Faraday's law: whenever the magnetic flux linking a conductor changes, a voltage is set up across the conductor. That voltage is called an induced e.m.f. (electromotive force), and its size is given by \[\varepsilon = -N\frac{\Delta\Phi}{\Delta t}\] where \(N\) is the number of turns and \(\Delta\Phi/\Delta t\) is the rate of change of magnetic flux. The minus sign is Lenz's law: the induced e.m.f. acts in the direction that opposes the change producing it.

The key distinction the question is testing is that induction produces a voltage, not a current, as its primary effect. A current only flows if that e.m.f. is connected to a complete circuit. This is why the e.m.f. still exists across the ends of a rod moved through a field even when the ends are not joined, and it is why a generator is rated by its e.m.f.

An eddy current is a circulating current, not a voltage; it is one of the consequences of induction inside a solid block of metal, and it is measured in amperes. Watts measure power, so that quantity cannot be a voltage at all. "Inductor voltage" is not a standard term in this topic; the recognised name for the quantity produced by a changing flux is the induced e.m.f. When a question names a unit or a quantity, check the dimensions first: only a quantity measured in volts can answer "a voltage called ...".

Bayanin Amsa

A thermometric property is any physical property that varies measurably, continuously and reproducibly with temperature, so that its value can be used as a scale of temperature. Different thermometers exploit different properties: a constant-volume gas thermometer uses pressure, a resistance thermometer uses electrical resistance, a thermocouple uses emf, and a liquid-in-glass thermometer uses the expansion of the liquid.

Mercury is used in liquid-in-glass thermometers, where the mercury is sealed in a bulb attached to a fine capillary tube. As the temperature rises the mercury expands, and because the bore is narrow a small increase in the volume of mercury produces a long, easily read movement of the thread. The property being used is therefore the change of volume with temperature, and mercury suits the job because it expands almost uniformly over a wide range (\(-39\,^\circ\text{C}\) to \(357\,^\circ\text{C}\)), is opaque and easily seen, is a good conductor of heat so it responds quickly, and does not wet glass.

Density does change with temperature, but only as a consequence of the volume change at fixed mass, and density is not what the instrument reads; the length of the mercury thread is a direct measure of volume. Pressure change belongs to gas thermometers, and resistance change belongs to platinum resistance thermometers, not to mercury in glass. When a question names a specific thermometric substance, identify the instrument it is used in first, because the instrument fixes which property is being measured.

Bayanin Amsa

For small oscillations a simple pendulum has period

and frequency is the reciprocal of period, \(f = 1/T\). Two preparation steps decide whether the arithmetic will be right: the length must be converted to metres, and the frequency must be taken at the end rather than confused with the period.

With \(L = 120\,\mathrm{cm} = 1.20\,\mathrm{m}\), \(g = 10\,\mathrm{m\,s^{-2}}\) and \(\pi = \frac{22}{7}\):

Hence

A useful sense check is that a pendulum about a metre long swings roughly once every two seconds, so its frequency must be about half a hertz. Any answer of a few hertz would mean several complete swings each second, which is physically impossible for a pendulum this long.

Two errors produce the other figures. Leaving the length as \(120\) instead of \(1.20\) inflates \(\sqrt{L/g}\) by a factor of about ten and drives the frequency badly wrong, and stopping at \(T\) and quoting \(2.2\) as though it were the frequency confuses seconds with hertz. Note also that the mass of the bob and the amplitude do not appear in the formula, so they never affect the answer for small swings.

Bayanin Amsa

Refractive index is always a ratio of two lengths measured in the same figure, and it is greater than one for light passing from air into glass. In the two standard constructions used with a glass block, the value is obtained as the larger measured length divided by the smaller:

• Snell construction: \(n = \dfrac{\sin i}{\sin r}\), where the two sines are represented by the perpendicular distances of two points, one on the incident ray and one on the refracted ray, from the normal at the point of incidence.
• Apparent-depth construction: \(n = \dfrac{\text{real depth}}{\text{apparent depth}}\), the real depth of a mark under the glass being the larger of the two.

With \(SQ = 10\,\mathrm{cm}\) and \(SR = 6\,\mathrm{cm}\), the ratio is

The value is dimensionless, which is why the centimetres cancel and no unit is quoted. It is also physically sensible: glass has a refractive index of about \(1.5\) to \(1.7\), and the corresponding speed of light in the glass would be \(v = c/n = 3.0\times10^{8}/1.67 = 1.8\times10^{8}\,\mathrm{m\,s^{-1}}\).

The most tempting wrong answer comes from inverting the ratio, \(6/10 = 0.60\). A refractive index less than one would mean light travels faster in the glass than in air, which cannot happen for light entering a denser medium; that value belongs to the reverse passage, glass to air, where \(n_{\text{glass}\to\text{air}} = 1/1.67 = 0.60\). Use this check every time: when light passes into the optically denser medium, divide so that the answer exceeds one, and remember that the ray bends towards the normal on entering the glass, so the angle in air is the larger one.

Bayanin Amsa

The critical angle \(C\) is the angle of incidence inside the denser medium at which the refracted ray just grazes along the boundary, so the angle of refraction in air is \(90^\circ\). Applying Snell's law at the glass-air boundary, \[n_{g}\sin C = n_{a}\sin 90^\circ.\] Taking \(n_a = 1\) for air and \(\sin 90^\circ = 1\), this rearranges to the standard result \[n = \frac{1}{\sin C}.\]

Substituting \(C = 45^\circ\), for which \(\sin 45^\circ = \dfrac{1}{\sqrt{2}}\): \[n = \frac{1}{1/\sqrt{2}} = \sqrt{2} \approx 1.41.\] The refractive index of the glass is \(\sqrt{2}\).

The frequent error is to write \(n = \sin C\), giving \(0.71\), a value less than one that would describe a medium in which light travels faster than in air. A refractive index for a denser medium relative to air is always greater than \(1\), so \(n = 1/\sin C\) is the correct arrangement, and \(\sin C\) small means \(n\) large. Keep the physical consequence in mind too: at any angle of incidence greater than \(45^\circ\) inside this glass, no light escapes and total internal reflection occurs, which is the principle behind optical fibres, prism periscopes and the sparkle of cut gemstones.

Bayanin Amsa

A radiation detector is any device that responds to the ionisation, excitation or chemical change produced when nuclear radiation passes through matter. To answer an "odd one out" question like this, check each device against that definition rather than against how familiar the name sounds.

An electrophorus is an electrostatic instrument. It is a flat insulating disc (or slab) with a metal plate and an insulating handle, used to produce charge repeatedly by friction and then by induction: the slab is charged by rubbing, the metal plate is placed on it and earthed briefly, and the plate carries away a charge of opposite sign. It measures nothing and detects nothing about radioactivity, so it is the device that does not belong in this list.

The other three are genuine detectors. A Geiger-Muller counter uses a gas-filled tube at high voltage in which an entering particle ionises the gas and triggers a pulse of current that is counted electronically. A scintillation counter or chamber uses a phosphor that emits a tiny flash of light when radiation strikes it; a photomultiplier converts each flash into an electrical pulse. A film badge contains photographic film that darkens in proportion to the dose received, so it records the total exposure of a worker over time. In the examination, sort nuclear-physics apparatus by the effect it exploits: ionisation of a gas, light emission, or blackening of photographic emulsion. Anything based only on charging by friction or induction belongs to electrostatics.

Bayanin Amsa

The temperature gradient of a lagged rod is the rate at which temperature falls along its length, defined as

Lagging matters because it stops heat escaping through the sides, so in the steady state the same heat flows through every cross-section and the temperature falls uniformly from one end to the other. That uniform fall is what makes a single gradient value meaningful.

Rearranging for the temperature difference across the whole rod:

A difference of \(83.2\ \text{K}\) is numerically the same as a difference of \(83.2\ ^\circ\text{C}\), because the kelvin and the Celsius degree are the same size; only the zeros of the two scales differ. Taking the given end as the cooler end, the other end is

so the temperature at the other end is about \(316\ ^\circ\text{C}\). The listed value of \(316.28\ ^\circ\text{C}\) is this result, the tiny difference in the final digit arising from rounding in the printed data.

Two points are worth noting. First, arithmetically the far end could also have been the cooler one, giving \(233-83.2 = 149.8\ ^\circ\text{C}\); that value is not among the choices, which fixes the given end as the cold end. Second, do not convert \(233\ ^\circ\text{C}\) to kelvin and then add the gradient result and forget to convert back, and do not multiply by the gradient without the length: \(83.2\) is a temperature difference, never a temperature. Always separate the difference calculation from the final scale reading.

Bayanin Amsa

An annular eclipse is a particular kind of solar eclipse. Like every solar eclipse it happens only when the sun, the moon and the earth lie on the same straight line with the moon in the middle, so that the moon's shadow falls on the earth. What makes it annular rather than total is the moon's distance: because the moon's orbit is elliptical, its angular size varies. When it is near its farthest point it appears slightly smaller than the sun, so the dark umbra does not quite reach the earth's surface and a bright ring (Latin annulus) of the sun's disc remains visible round the black moon.

Among the statements offered, the one that correctly describes the condition for this event is that the sun, moon and earth come into a straight line. That alignment, called syzygy, is the necessary geometry for both the total and the annular solar eclipse; the difference between them is only the apparent size of the moon at the time.

The statement that the earth comes between the moon and the sun describes a lunar eclipse, in which the earth's shadow falls on the moon; that is the commonest confusion in this topic, so fix the order clearly: in a solar eclipse the moon blocks the sun from the earth, in a lunar eclipse the earth blocks the sun from the moon. A gathering of stars is a cluster or constellation and has nothing to do with eclipses, and simple invisibility of one body is not a definition of an eclipse, since the moon is invisible at new moon in every month without any eclipse occurring. In the examination, first identify which body is being shadowed, then decide whether the shadow is total, partial or annular.

Bayanin Amsa

Volume expansion is governed by the cubic expansivity \(\gamma\): \[\Delta V = V_1 \gamma\, \Delta\theta.\] For a solid the three expansivities are related by \(\gamma = 3\alpha\) and \(\beta = 2\alpha\), because a solid expands by the same fractional amount in each of its three perpendicular directions. Here the linear expansivity is given, so convert first: \[\gamma = 3\alpha = 3 \times 2.0\times10^{-5} = 6.0\times10^{-5}\,\text{K}^{-1}.\]

Now substitute the data, with \(V_1 = 1\,\text{cm}^3\) and \(\Delta V = 0.0018\,\text{cm}^3\): \[0.0018 = 1 \times 6.0\times10^{-5} \times \theta \quad\Rightarrow\quad \theta = \frac{0.0018}{6.0\times10^{-5}} = 30.\] The temperature rise is \(30\) kelvin, which is a rise of \(30\,^\circ\text{C}\). A change of temperature has the same numerical value on both scales because the degree sizes are identical, which is why an expansivity quoted in \(\text{K}^{-1}\) may be used directly with a Celsius temperature change.

The mistake that produces \(90\) is using \(\alpha\) itself in the volume formula, and the mistake that produces \(45\) is using \(\beta = 2\alpha\), the area expansivity. Match the expansivity to the dimension being measured: length with \(\alpha\), area with \(2\alpha\), volume with \(3\alpha\). Note also that only the ratio \(\Delta V / V_1\) matters, so the units of volume cancel and no conversion of cubic centimetres is needed.

Bayanin Amsa

The SI system is built on seven base (fundamental) units which are defined independently of one another: the metre, kilogram, second, ampere, kelvin, mole and candela. Every other unit is a derived unit, meaning it can be written as a combination of these base units. So the task here is simply to test each unit for whether it can be broken down further.

The ampere is the base unit of electric current, so it cannot be expressed in terms of anything more fundamental. The other three all reduce to combinations of base units:

• joule: work \(=\) force \(\times\) distance, so \(1\,\mathrm{J} = 1\,\mathrm{N\,m} = 1\,\mathrm{kg\,m^{2}\,s^{-2}}\);
• volt: \(V = W/Q\), so \(1\,\mathrm{V} = 1\,\mathrm{J\,C^{-1}} = 1\,\mathrm{kg\,m^{2}\,s^{-3}\,A^{-1}}\);
• ohm: \(R = V/I\), so \(1\,\Omega = 1\,\mathrm{kg\,m^{2}\,s^{-3}\,A^{-2}}\).

A common misconception is that any unit with its own special name, such as the joule or the volt, must be fundamental. The special name is only a convenience; what matters is whether the unit can be written in terms of others. Notice too that the coulomb is not a base unit even though charge feels more basic than current: the SI system defines the ampere first and then treats \(1\,\mathrm{C} = 1\,\mathrm{A\,s}\). Memorise the seven base units and their quantities, then any question of this type becomes a single-step elimination.

Bayanin Amsa

Doping means adding a controlled trace of impurity to a pure semiconductor such as silicon or germanium to increase the number of mobile charge carriers. Silicon has four valence electrons and forms four covalent bonds.

• Adding a pentavalent impurity (phosphorus, arsenic, antimony) supplies five valence electrons per atom. Four complete the bonds and the fifth is left loosely bound, becoming a free electron. The material is n-type and electrons are its majority carriers.
• Adding a trivalent impurity (boron, aluminium, gallium) supplies only three valence electrons, so one bond is left short of an electron. That vacancy is a hole, and it moves through the lattice as neighbouring electrons hop into it, behaving like a mobile positive charge of magnitude \(+e\). The material is p-type and holes are its majority carriers.

Both kinds of carrier are present in any doped sample, one as the majority and the other as the minority produced by thermal generation, so the charge carriers in doped semiconductors are electrons and holes.

The distractors rest on real misconceptions. Protons and neutrons are locked in the nuclei of the fixed lattice atoms and cannot migrate, so they never carry current in a solid. Anions and cations do carry charge, but that is electrolytic conduction in a solution or molten salt, where whole ions drift; a semiconductor crystal keeps its atoms in place and moves only electrons and the holes they leave behind. Remember for the examination that conventional current in a p-type region is described as a flow of holes in the direction of the field, while the electrons that actually move travel the opposite way.

Bayanin Amsa

All electromagnetic waves travel at the same speed \(c = 3.0\times 10^{8}\ \text{m s}^{-1}\) in a vacuum, and they satisfy \(c = f\lambda\). Since \(c\) is fixed, wavelength and frequency are inversely related: the shortest wavelength belongs to the highest frequency, and therefore to the most energetic radiation, because \(E = hf\).

Ordering the members of the spectrum given here from long wavelength to short: infrared, then visible light, then ultraviolet, then X-rays. Of these, X-rays have the shortest wavelength, of the order of \(10^{-10}\ \text{m}\), compared with about \(10^{-8}\ \text{m}\) for ultraviolet, \(4\times 10^{-7}\) to \(7\times 10^{-7}\ \text{m}\) for visible light and around \(10^{-5}\ \text{m}\) for infrared. The table below sets out the comparison.

Ultraviolet is the tempting alternative because it is the one most students associate with harmful, penetrating radiation from the Sun, but it sits between visible light and X-rays. The very short wavelength of X-rays is precisely why they penetrate soft tissue and are diffracted by the regular spacing of atoms in crystals, an effect that only works when the wavelength is comparable with atomic spacing. A reliable method in the examination is to recite the spectrum in a fixed order, from radio waves through microwaves, infrared, visible light, ultraviolet and X-rays to gamma rays, remembering that wavelength decreases and frequency increases along that sequence, then read off whichever end the question asks for.

Bayanin Amsa

Newton's law of universal gravitation states that the attractive force between two point masses is

where \(G\) is the universal gravitational constant and \(r\) is the distance between their centres. The force is therefore directly proportional to the product of the two masses, and this question asks only how that product changes.

Doubling each mass replaces \(m_1 m_2\) by \((2m_1)(2m_2) = 4m_1 m_2\), while \(r\) is unchanged. Writing the new force as \(F_2\) and dividing one expression by the other lets \(G\) and \(r\) cancel:

The mistake to guard against is doubling the force to \(20\,\mathrm{N}\), which comes from doubling only one mass, or from treating the force as proportional to the sum of the masses rather than their product. A second useful habit for this formula is to keep the two dependences separate: the force scales with each mass to the first power but with distance to the power \(-2\). So if the masses were doubled and the separation also doubled, the factor would be \(4 \times \tfrac{1}{4} = 1\) and the force would stay at \(10\,\mathrm{N}\). Setting up the ratio \(F_2/F_1\) rather than trying to find \(G\) or the actual masses is always the fastest and safest method in these proportionality questions.

Bayanin Amsa

This question uses de Broglie's idea that a moving particle has a wavelength linked to its momentum: \[\lambda = \frac{h}{p} = \frac{h}{mv},\] so that \[m = \frac{h}{\lambda v}.\] Everything therefore depends on getting the wavelength into metres, because \(h\) is in \(\text{J s}\) and the speed in \(\text{m s}^{-1}\).

The wavelength is given in millimetres, so convert first: \[\lambda = 4.0 \times 10^{-7}\ \text{mm} = 4.0 \times 10^{-7} \times 10^{-3}\ \text{m} = 4.0 \times 10^{-10}\ \text{m}.\] Now substitute: \[m = \frac{6.63 \times 10^{-34}}{(4.0 \times 10^{-10})(2.7 \times 10^{8})} = \frac{6.63 \times 10^{-34}}{1.08 \times 10^{-1}} = 6.1 \times 10^{-33}\ \text{kg}.\] The significant figures come out as 6.1, so the intended choice is the value quoted with those figures; its power of ten appears to be misprinted, since the correct working gives \(6.1 \times 10^{-33}\ \text{kg}\) rather than \(10^{-31}\). Quote \(6.1 \times 10^{-33}\ \text{kg}\) as your worked answer and select the value beginning 6.1.

The step that costs most marks is the millimetre-to-metre conversion. Skipping it, and using \(4.0 \times 10^{-7}\ \text{m}\), gives \(6.1 \times 10^{-36}\ \text{kg}\), a thousand times too small. A second slip is inverting the relation and multiplying by \(\lambda v\) instead of dividing. As a check on the physics, remember the inverse proportionality: a shorter wavelength means a larger momentum, so a heavier or faster particle always has the smaller de Broglie wavelength, which is why wave behaviour is only observed for very light particles such as electrons.

Bayanin Amsa

Myopia (short-sightedness) is a defect of vision in which distant objects cannot be seen clearly because the eye focuses light in front of the retina. The far point (the farthest distance at which objects are seen clearly) is closer than infinity - in this case, 95 cm.

To correct myopia, a diverging (concave) lens is placed before the eye. The lens diverges incoming parallel rays from distant objects so that they appear to come from the person's far point, which the eye can then focus on the retina.

The required focal length of the correcting lens equals the far point distance. Since the lens must produce a virtual image at 95 cm for an object at infinity:

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-95} - \frac{1}{\infty} = -\frac{1}{95} \]

So \( f = -95 \text{ cm} \) (negative sign confirms a diverging lens).

The correction is a diverging lens of focal length 95 cm. A converging lens would worsen myopia, and a mirror is not used to correct refractive eye defects.

Bayanin Amsa

A hydraulic press works by Pascal's principle: pressure applied to an enclosed incompressible liquid is transmitted equally throughout, so the pressure under the small piston equals the pressure under the large piston. \[\frac{F_1}{A_1} = \frac{F_2}{A_2}.\] Since each piston is circular, \(A = \pi r^2\), and the \(\pi\) cancels: \[\frac{F_1}{r_1^{2}} = \frac{F_2}{r_2^{2}} \quad\Rightarrow\quad \frac{r_2^{2}}{r_1^{2}} = \frac{F_2}{F_1}.\]

Substituting the given forces, \[\frac{r_2^{2}}{r_1^{2}} = \frac{3200}{200} = 16 \quad\Rightarrow\quad \frac{r_2}{r_1} = \sqrt{16} = 4.\] So \(r_1 : r_2 = 1 : 4\).

The decisive step is the square root. Forces in a hydraulic press scale with area, and area scales with the square of the radius, so a force multiplication of \(16\) needs a radius ratio of only \(4\), not \(16\). Reading off \(1:16\) is the classic error, made by matching the force ratio straight to the radii; \(1:2\) comes from taking the square root twice.

Remember also that the press multiplies force but not energy: the small piston must travel \(16\) times as far as the large one, since the same volume of liquid is displaced, \(A_1 d_1 = A_2 d_2\). In an examination, decide first whether the ratio you are asked for is one of areas, radii or diameters, and insert or remove the square accordingly.

Bayanin Amsa

This is a method-of-mixtures problem, and the governing statement is the principle of conservation of energy: with the container ignored and no loss to the surroundings, \[\text{heat lost by the hot water} = \text{heat gained by the cold water}.\] Each term is calculated from \(Q = mc\,\Delta\theta\). Boiled water is at \(100\,^\circ\text{C}\), and the final mixture temperature is \(60\,^\circ\text{C}\), so the temperature changes are:

• hot water cools from \(100\,^\circ\text{C}\) to \(60\,^\circ\text{C}\): \(\Delta\theta = 40\,\text{K}\)
• cold water warms from \(25\,^\circ\text{C}\) to \(60\,^\circ\text{C}\): \(\Delta\theta = 35\,\text{K}\)

Both liquids are water, so the specific heat capacity \(c\) is the same on each side and cancels: \[m \times c \times 40 = 8 \times c \times 35\] \[40m = 280 \quad\Rightarrow\quad m = 7\,\text{kg}.\] Seven kilograms of boiled water is required.

Three points decide this question. First, "boiled water" fixes the hot temperature at \(100\,^\circ\text{C}\); it is data given in words rather than symbols. Second, the two temperature changes are different (\(40\,\text{K}\) against \(35\,\text{K}\)), so the masses cannot simply be equal, and the hot mass must be the smaller multiple: \(m/8 = 35/40\). Third, because both substances are water, \(c\) never needs a numerical value, so quoting \(4200\,\text{J kg}^{-1}\text{K}^{-1}\) adds arithmetic but no information. Also note that no latent heat appears here: nothing changes state, the steam having already condensed. In an examination, write out both \(\Delta\theta\) values explicitly before forming the equation, since reversing them is the commonest source of a wrong mass.

Bayanin Amsa

Dimensions describe a quantity in terms of the base quantities mass \(M\), length \(L\) and time \(T\), independent of the units used. The key physical insight here is that tension in a string is simply the force the string exerts along its length. It is not a special new quantity, so it must have exactly the dimensions of force.

Get those dimensions from Newton's second law, \(F = ma\). Mass contributes \(M\). Acceleration is velocity change per unit time, that is \(\frac{L\,T^{-1}}{T} = L\,T^{-2}\). Multiplying: \[[F] = M \times L\,T^{-2} = M\,L\,T^{-2}.\] So the dimensional formula of tension is \(M\,L\,T^{-2}\), whose SI unit, the newton, is correspondingly \(\text{kg}\,\text{m}\,\text{s}^{-2}\).

Watch the sign of the time index. Writing \(M\,L\,T^{2}\) would mean force grows with the square of time, which is dimensionally the same as mass times length times time squared and matches no mechanical quantity here; the index is negative because time appears in the denominator of acceleration twice. An expression with no \(M\) at all, such as \(L\,T^{-2}\), is the dimension of acceleration alone, not of a force, and raising \(M\) to a power other than one has no justification since force is directly proportional to a single mass. Exam takeaway: whenever a question asks for the dimensions of tension, thrust, weight, upthrust or any pull or push, answer with the dimensions of force, \(M\,L\,T^{-2}\).

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The power of a lens measures how strongly it converges or diverges light. A lens that bends rays sharply brings them to a focus close to the lens, so it has a short focal length; a weak lens focuses rays far away. Power is therefore defined as the reciprocal of the focal length, \[P = \frac{1}{f},\] with \(f\) in metres. The unit of \(P\) is the dioptre (\(\text{D}\)), which is simply \(\text{m}^{-1}\). So the power in dioptres is \(\frac{1}{f}\).

Two details make the definition work. First, \(f\) must be expressed in metres before taking the reciprocal: a lens of focal length \(20\,\text{cm} = 0.20\,\text{m}\) has \[P = \frac{1}{0.20} = +5.0\,\text{D}.\] Second, the sign of \(f\) carries through, so a converging (convex) lens has positive power and a diverging (concave) lens has negative power. Powers also add for thin lenses placed in contact, \(P = P_1 + P_2\), which is exactly why opticians quote lenses in dioptres rather than in centimetres.

Expressions such as \(f\), \(2f\) or \(3f\) cannot be correct because they grow as the focal length grows, which would say that a lens focusing light far away is the more powerful one. They also have the wrong unit: metres instead of \(\text{m}^{-1}\). A quick unit check on any formula offered in an optics question will usually eliminate the distractors immediately, and remember to convert centimetres to metres before computing a dioptre value.

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This question links the electrical energy supplied by the kettle to the heat energy gained by the water. Assuming no heat is lost, the electrical energy delivered in time \(t\) equals the heat needed to raise the water's temperature:

Work out each side separately. The heat required is

The power of the kettle is

Since power is energy per second, the time taken is

Two slips account for the other figures. Using the final temperature \(65\ ^\circ\text{C}\) instead of the temperature rise of \(35\ \text{K}\) inflates the energy badly, and halving or doubling the power (for instance by dividing by \(2400\) instead of \(1200\)) gives \(245\ \text{s}\), which is the trap set here. Also note that a temperature change of \(35\ ^\circ\text{C}\) is numerically identical to \(35\ \text{K}\), so the specific heat capacity in \(\text{J kg}^{-1}\text{K}^{-1}\) can be used directly without converting to kelvin. Always compute the temperature difference first and write it down before substituting.

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The thermal capacity (heat capacity) of a body is the quantity of heat needed to raise the temperature of the whole body by one kelvin, measured in \(\text{J K}^{-1}\). It is related to the specific heat capacity \(c\) by

Reading that equation tells you exactly what \(C\) depends on. It depends on the mass \(m\) of the body, and on \(c\), which is fixed by the substance the body is made of, that is by its nature or material. So thermal capacity depends on the mass and the nature of the body, and on nothing else.

The quantity of heat supplied is not a factor, because \(C\) is a ratio, \(C = Q/\Delta\theta\); supplying twice the heat produces twice the temperature rise and leaves \(C\) unchanged. Temperature is not a factor either: \(C\) tells you how much heat is needed per kelvin, whichever kelvin you start from, so a body at \(20\ ^\circ\text{C}\) and the same body at \(80\ ^\circ\text{C}\) have essentially the same thermal capacity. Volume is not an independent factor because, for a given material, volume is only another way of stating mass through the density, \(m = \rho V\); once the mass and the material are named, the volume adds nothing.

A concrete check makes this memorable. Two blocks of the same mass, one aluminium and one lead, need very different amounts of heat for the same rise, which shows the nature matters, and two aluminium blocks of different masses also need different amounts, which shows the mass matters. Distinguish carefully in the examination: specific heat capacity \(c\), in \(\text{J kg}^{-1}\text{K}^{-1}\), depends only on the nature of the substance, while thermal capacity \(C\), in \(\text{J K}^{-1}\), depends on the nature and on how much of it there is.

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A standing (stationary) wave is not a wave that travels; it is the pattern formed when two identical progressive waves of the same frequency and amplitude travel through the same region in opposite directions and superpose. In practice the second wave is supplied by reflection: a wave sent along a stretched string or down a pipe bounces back from the fixed end or the closed end and overlaps the incoming wave. So a standing wave is produced when a wave reflects off a boundary and interferes with itself.

Where the two waves always arrive in step, constructive interference gives points of maximum displacement called antinodes; where they always arrive exactly out of step, destructive interference gives points of permanently zero displacement called nodes. Because the nodes and antinodes stay in fixed positions, no energy is carried along the medium, which is exactly what distinguishes a standing wave from a progressive one. This is why a guitar string, an organ pipe and a microwave oven cavity all show fixed loud and quiet or bright and dark positions.

The alternatives describe different physics. A wave vibrating in a vertical plane is simply a plane-polarised transverse wave, and the word "standing" in the term refers to the pattern not moving along the medium, not to the direction of vibration. Motion of the source towards or away from the observer changes the observed frequency and is the Doppler effect, which involves a single travelling wave and no superposition at all. Exam reminder: link standing waves to the two conditions of reflection and superposition, and to the presence of fixed nodes and antinodes.

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This is a vector-addition problem, so the two journeys must be resolved into perpendicular components before they are combined. The bearing notation \(N30^\circ E\) means the direction is measured \(30^\circ\) away from north, turning towards the east. For a displacement of \(10.0\,\text{m}\) in that direction, north is the adjacent side and east the opposite side of the \(30^\circ\) angle:

• eastward part \(= 10.0\sin 30^\circ = 5.0\,\text{m}\)
• northward part \(= 10.0\cos 30^\circ = 8.66\,\text{m}\)

The first leg is entirely eastward, so the totals are \[x = 6.0 + 5.0 = 11.0\,\text{m (east)},\qquad y = 0 + 8.66 = 8.66\,\text{m (north)}.\] These two totals are at right angles, so Pythagoras gives the straight-line distance from the start: \[r = \sqrt{11.0^2 + 8.66^2} = \sqrt{121 + 75.0} = \sqrt{196} = 14.0\,\text{m}.\] The man is \(14.0\,\text{m}\) from his starting point.

The usual error is to add the magnitudes, \(6.0 + 10.0 = 16.0\,\text{m}\), or to interchange the sine and cosine because the angle was assumed to be measured from the east line. In bearings written as \(N\theta E\) the angle is measured from north, so north takes the cosine. Sketching the two arrows head-to-tail, as above, shows at once which component belongs to which trigonometric ratio.

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Waves divide into two families. Mechanical waves, such as water waves, sound and waves on a string, are oscillations of the particles of a material medium, so they cannot exist without that medium. Electromagnetic waves, such as light, radio waves and X-rays, are oscillations of electric and magnetic fields, which need no particles at all and therefore travel through a vacuum at \(3.0\times10^{8}\,\mathrm{m\,s^{-1}}\). This is the general difference between water waves and light waves: the medium of propagation each requires.

The evidence for it is everyday. Sunlight reaches the earth across the emptiness of space, whereas a water wave dies out the moment the water ends at a shoreline, and a ripple tank produces no waves when the water is drained. This is also why light from distant stars reaches us but their sound never does.

The other suggested differences do not hold. Both kinds of wave can be reflected, water waves from a barrier in a ripple tank and light from a mirror, and both can be diffracted, water waves spreading through a narrow gap between barriers and light spreading at the edge of an obstacle or through a fine slit. The direction of vibration is not a general point of difference either, because water surface waves and light waves are both transverse: the displacement is perpendicular to the direction of travel in each case. In the examination, when asked to distinguish two waves, first classify each as mechanical or electromagnetic, since that single classification decides the need for a medium, the possible speeds, and whether the wave can be polarised.

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This question tests Charles' law: for a fixed mass of gas at constant pressure, the volume is directly proportional to the absolute (kelvin) temperature, so \(\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\).

The temperatures must be converted to kelvin before they are substituted, because the proportionality only holds on a scale whose zero is absolute zero:

• \(T_1 = 0 + 273 = 273\ \text{K}\)
• \(T_2 = 273 + 273 = 546\ \text{K}\)

Substituting,

The absolute temperature doubles, so the volume doubles to \(400\ \text{m}^3\).

Two mistakes are common. The first is using the Celsius values directly, which produces the meaningless ratio \(273/0\) and tempts a student into a wrong figure. The second is assuming that because the mass is fixed the volume cannot change; a fixed mass only means no gas enters or leaves, and the gas is still free to expand. A volume of \(200\ \text{m}^3\) would require the temperature to be unchanged, and \(100\ \text{m}^3\) would require the absolute temperature to be halved, neither of which happens here. In every gas-law calculation, convert to kelvin as the very first step.

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This question tests magnetic permeability, which is a measure of how easily a material allows magnetic field lines to pass through it. Magnetic shielding does not work by blocking field lines, because magnetic field lines cannot simply be stopped. It works by offering the field lines a much easier path that carries them around the region you want to protect.

Soft iron has a very high relative permeability, several thousand times that of air. When an instrument is enclosed in a soft iron case, nearly all of the external field lines are pulled into the iron walls and guided around the cavity, leaving the space inside with an extremely weak field. Soft iron rather than steel is used because soft iron has low retentivity: it magnetises strongly while the external field is present, but loses almost all of that magnetism once the field is removed, so the screen itself does not become a permanent magnet that would disturb the instrument.

Aluminium, brass and copper are non-magnetic. Their relative permeability is essentially the same as that of air, so field lines pass straight through them and the enclosed region is not protected. Copper and aluminium do oppose a changing magnetic field through induced eddy currents, which is why they appear in electrical screening, but against a steady magnetic field they provide no shielding. A useful examination link is: magnetic screening requires high permeability with low retentivity, and that combination describes soft iron.

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The diagram shows a gold-leaf electroscope that is already positively charged, as indicated by the diverged leaves marked with positive (+) signs. When a positively charged rod is brought near the cap, electrostatic induction occurs.

Since the electroscope already carries a net positive charge, the approaching positive rod repels additional positive charges from the cap region down through the stem and onto the leaves. This increases the concentration of positive charge on both leaves, causing the electrostatic repulsion between them to grow stronger.

As a result, the leaves spread further apart and the divergence increases. This is a standard demonstration of charge interaction: like charges repel, and adding more of the same sign of charge to the leaves amplifies their mutual repulsion.

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When two surfaces are in contact, the contact force between them can be resolved into two parts. The component perpendicular (normal) to the surface is the normal reaction, and the component along the surface, that is tangential to it, is friction. The definition given in the question specifies a force that is tangential to the surface and that opposes sliding, and that is precisely the frictional force.

Friction arises from the interlocking of microscopic irregularities and from attraction between the molecules of the two surfaces at the points where they genuinely touch. It always acts along the surface and always in the direction that opposes relative sliding, or the tendency to slide, which is why a stationary block on a rough incline does not slip and why a pushed box eventually stops. For a body on the point of sliding, \(F = \mu R\), where \(R\) is the normal reaction and \(\mu\) the coefficient of friction, a relation which itself shows that friction and the normal reaction are two distinct, mutually perpendicular quantities.

The normal force is the tempting alternative, but it acts at right angles to the surface and pushes the body away from the surface; it supports the body rather than resisting its sliding. Weight, \(W = mg\), is the gravitational pull of the Earth and acts vertically downwards regardless of any surface, so it is not tangential except in the special case of a vertical wall. Upthrust is the upward force a fluid exerts on a body immersed in it, again vertical and not a surface-contact tangential force. In the examination, use the direction words as your key: perpendicular to the surface means normal reaction, along the surface means friction.

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All the relationships follow from the ideal gas equation \[PV = nRT.\] To decide whether two quantities are directly or inversely proportional, hold the other two constant and see what the equation demands.

Only pressure and volume sit on the same side of the equation as a product, and a product held constant is the definition of inverse proportionality. So the inversely proportional pair is pressure and volume: squeeze a fixed mass of gas at constant temperature into half the space and the pressure doubles, because the molecules strike the walls twice as often.

A practical way to confirm the type of proportionality is the shape of the graph. Pressure against volume gives a curve (a hyperbola), while pressure against \(1/V\) gives a straight line through the origin. Pressure against absolute temperature and volume against absolute temperature both give straight lines through the origin directly. In an examination, always state which quantities are being held constant before quoting a gas law, since the same two variables can behave differently if a third is allowed to vary.

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This question is about streamlining. When a body moves through a fluid such as air or water, the fluid must be pushed aside and made to flow round the body. A blunt front forces the fluid to change direction abruptly, the flow behind it breaks up into swirling eddies, and the pressure in front becomes much higher than the pressure behind. That pressure difference, together with the rubbing of the fluid layers along the surface, makes up the resistive force called drag or fluid friction.

A pointed, tapered front lets the fluid part smoothly and rejoin gradually behind the body, so the flow stays streamlined instead of turbulent and the pressure difference between front and back is much smaller. The result is a reduction in the fluid friction acting on the boat or aircraft, which means less driving force is needed for a given speed, less fuel is used, and a higher top speed becomes possible for the same engine power. This is why fast-moving objects in nature and in engineering, from fish and birds to aircraft and racing hulls, all share the same tapered shape.

The suggestion that the shape increases fluid friction reverses the physics: increasing drag would waste energy, and shapes deliberately made blunt, such as a parachute canopy, are used precisely when large drag is wanted. Stopping depends on reverse thrust, brakes or drag devices, not on the shape of the nose, and appearance is not a physical explanation. In the examination, treat any question about the shape of a moving vehicle as a question about minimising drag, and be ready to name the mechanism as smooth, streamlined flow replacing turbulent flow.

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The circuit diagram shows a 16 V battery connected to a single 2 Ω resistor in a closed loop. To find the power dissipated in the resistor, use the formula:

\(P = \frac{V^2}{R}\)

Substituting the values:

\(P = \frac{(16)^2}{2} = \frac{256}{2} = 128 \text{ W}\)

The total power dissipated in the circuit is 128 W.

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Electrical power delivered to a device is the product of the potential difference across it and the current through it: \[P = IV.\] The rating on an appliance states the power it consumes at its working voltage, so the current follows by rearranging: \[I = \frac{P}{V} = \frac{500}{220} = 2.27\,\text{A}\ (3\ \text{s.f.}).\] The oven therefore draws about \(2.27\,\text{A}\).

It is worth seeing where the other numbers could come from, because each represents a specific error. Dividing the voltage by the power, \(220/500\), gives \(0.44\), while using a mains value of \(110\,\text{V}\) instead of \(220\,\text{V}\) would double the answer to \(4.55\,\text{A}\). Only the direct substitution into \(I = P/V\) with the values actually given is defensible.

Two related results are often needed in the same question and follow from the same data: the resistance of the heating element at working temperature is \[R = \frac{V^2}{P} = \frac{220^2}{500} = 96.8\,\Omega,\] and the energy consumed in, for example, half an hour is \[E = Pt = 500 \times 1800 = 9.0 \times 10^{5}\,\text{J} = 0.25\,\text{kWh}.\] Keep the three forms \(P = IV = I^2R = \dfrac{V^2}{R}\) at hand, and choose the one whose quantities are actually given rather than working through an intermediate you do not need.

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Specific heat capacity is the heat needed to raise the temperature of one kilogram of a substance by one kelvin. It comes from the heat equation \[Q = mc\Delta\theta,\] where \(Q\) is the heat supplied in joules, \(m\) the mass in kilograms and \(\Delta\theta\) the temperature rise. Because heat loss to the surroundings is stated to be negligible, all 1000 J supplied goes into the rod, so no correction is needed.

Making \(c\) the subject and substituting: \[c = \frac{Q}{m\Delta\theta} = \frac{1000}{0.025 \times 15} = \frac{1000}{0.375} = 2666.7\ \text{J kg}^{-1}\text{K}^{-1}.\] So the specific heat capacity is \(2666.7\ \text{J kg}^{-1}\text{K}^{-1}\).

Note that the temperature rise needs no conversion. A change of \(15\ ^\circ\text{C}\) is a change of 15 K because the two scales have the same size of degree, so adding 273 here is a wasted step that produces a badly wrong answer. The other frequent slip is working out the denominator carelessly: \(0.025 \times 15 = 0.375\), not 0.0375 or 3.75. Exam reminder: distinguish specific heat capacity \(c\), measured in \(\text{J kg}^{-1}\text{K}^{-1}\), from heat capacity \(C = mc\), measured in \(\text{J K}^{-1}\); the units in the options tell you which one is wanted.

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When a solid melts at its melting point, the heat supplied is used to break down the rigid arrangement of the particles rather than to raise the temperature, so a thermometer in the mixture stays at \(0\ ^\circ\text{C}\) throughout. Heat that produces a change of state at constant temperature is called latent heat, the word latent meaning hidden, because it produces no temperature reading.

The distinction the question turns on is between a total quantity and a per-kilogram quantity. The specific latent heat of fusion \(l\) is the heat needed to melt one kilogram of the solid at its melting point, with the unit \(\text{J kg}^{-1}\). The latent heat of fusion is the heat needed to melt the whole given mass, so

with the unit joule. Because the question fixes a definite mass of \(5\ \text{kg}\), the quantity described is the latent heat of fusion of that ice, not the specific latent heat. Naming it as the specific quantity would be wrong by a factor of \(5\).

The heat-capacity terms do not apply at all, because both describe heat that causes a temperature change: heat capacity is \(Q/\Delta\theta\) in \(\text{J K}^{-1}\) and specific heat capacity is \(Q/(m\Delta\theta)\) in \(\text{J kg}^{-1}\text{K}^{-1}\). Here the temperature does not change, so any formula containing \(\Delta\theta\) is ruled out immediately.

Carry two habits into the examination. First, the word specific always means per unit mass, so it can only be used when no particular mass is mentioned. Second, decide whether the heat causes a temperature change or a change of state: use \(Q = mc\Delta\theta\) for the first and \(Q = ml\) for the second.